234. Palindrome Linked List(判断链表是否回文)

Given a singly linked list, determine if it is a palindrome.

Follow up:
Could you do it in O(n) time and O(1) space?

思路：

1、 利用快慢指针找到链表中点

2、从中点开始反转链表，判断是否相等。

class Solution {
    public boolean isPalindrome(ListNode head) {
        ListNode faster= head ,slower= head;
        while(faster!=null && faster.next!=null){
            faster = faster.next.next;
            slower = slower.next;
        }
        if(faster!=null)
            //奇数
            slower=slower.next;
        ListNode newhead = Reverse(slower);
        while(newhead!=null){
            if(head.val!=newhead.val){
                return false;
            }
            head = head.next;
            newhead = newhead.next;
        }
            return true;

    }

    public ListNode  Reverse(ListNode head) {
        if(head == null ||head.next == null) return head;
        ListNode pre = head;
        head = head.next;
        pre.next = null;
        while(head != null){
            ListNode next = head.next;
            head.next = pre;
            pre = head;
            head = next;
        }
        return pre;
    }

}